Question

how to write the equation of projectile after it has reached its maximum height(suppose at time t1 and t2 there are at same height) without finding the the time taken to reach at it full height

Answer 1

@Michele_Laino

Answer 2

@Michele_Laino pls help in hurry i want to solve this doubt

Answer 3

is there any way

Answer 4

in general I solve that type of problem using vectors equations, namely:
\[\large \left\{ \begin{gathered}
{\mathbf{OP}}\left( t \right) = {\mathbf{O}}{{\mathbf{P}}_{\mathbf{0}}} + {{\mathbf{v}}_{\mathbf{0}}}t + \frac{1}{2}{\mathbf{g}}{t^2} \hfill \\
{\mathbf{v}}\left( t \right) = {{\mathbf{v}}_{\mathbf{0}}} + {\mathbf{g}}t \hfill \\
\end{gathered} \right.\]

Answer 5

Then I consider a reference frame and I will rewrite those equations in that reference frame

Answer 6

for example, if we pick the subsequent reference frame:

Answer 7

then those vector equation can be rewritten as follows:
\[\Large \begin{gathered}
\left\{ \begin{gathered}
z\left( t \right) = {z_0} + {v_0}\left( {\sin \theta } \right)t - \frac{1}{2}g{t^2} \hfill \\
x\left( t \right) = {x_0} + {v_0}\left( {\cos \theta } \right)t \hfill \\
\end{gathered} \right. \hfill \\
\hfill \\
\left\{ \begin{gathered}
{v_z}\left( t \right) = {v_0}\left( {\sin \theta } \right) - gt \hfill \\
{v_x}\left( t \right) = {v_0}\left( {\cos \theta } \right) \hfill \\
\end{gathered} \right. \hfill \\
\end{gathered} \]